Transcript Lecture 13: Inverse Laplace Transform 5 Laplace transform (3 lectures): Laplace transform as Fourier transform with convergence factor.

Lecture 13: Inverse Laplace Transform
5 Laplace transform (3 lectures):
Laplace transform as Fourier transform with
convergence factor. Properties of the Laplace
transform
Specific objectives for today:
• Poles and zeros of a Laplace transfer function
• Rational polynomial transfer functions
• Inverse Laplace transform
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Lecture 13: Resources
Core material
SaS, O&W, Chapter 9.2(end), 9.3, 9.4
Recommended material
MIT, Lecture 17, 18
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Reminder: Laplace Transforms
Equivalent to the Fourier transform when s=jw

X (s)   x(t )est dt

L
x(t )  X ( s )
There is an associated region of convergence for s
where the (transformed) signal has finite energy. The
Laplace transform is only defined for these values
Laplace transform is linear (easy!)
Examples for the Laplace transforms include
1
, Re{ s}   a
sa
L
s 1
 2t
t
3e u (t )  2e u (t )  2
,
s 3s  2
 at
L
e u (t ) 
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Re{ s}  1
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Ratio of Polynomials
In each of these examples, the Laplace transform is rational, i.e. it is a
ratio of polynomials in the complex variable s.
N ( s)
X ( s) 
D( s )
where N and D are the numerator and denominator polynomial
respectively.
In fact, X(s) will be rational whenever x(t) is a linear combination of
real or complex exponentials. Rational transforms also arise
when we consider LTI systems specified in terms of linear,
constant coefficient differential equations.
We can mark the roots of N and D in the s-plane along with the ROC
Example 3:
Im
s-plane
– roots of N(s)
-2
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x
x
-1
1
Re
x – roots of D(s)
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Poles and Zeros
The roots of N(s) are known as the zeros. For these values of s,
X(s) is zero.
The roots of D(s) are known as the poles. For these values of s,
X(s) is infinite, the Region of Convergence for the Laplace
transform cannot contain any poles, because the corresponding
integral is infinite
The set of poles and zeros completely characterise X(s) to within a
scale factor (+ ROC for Laplace transform)
 ( s  zi )
X (s)  i
 j (s  p j )
The graphical representation of X(s) through its poles and zeros in
the s-plane is referred to as the pole-zero plot of X(s)
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Example: Poles and Zeros
Consider the signal:
4
1
x(t )   (t )  e t u (t )  e 2t u (t )
3
3
By linearity (& last lecture) we can evaluate the second and third
terms
The Laplace transform of the impulse function is:

L{ (t )}    (t )e st dt  1

which is valid for any s. Therefore,
4 1
1 1
X ( s)  1 

3 s 1 3 s  2
( s  1) 2

,
Re{s}  2
( s  1)(s  2)
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Im
-1
x
x
1
2
Re
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ROC Properties for Laplace Transform
Property 1: The ROC of X(s) consists of strips parallel to the
jw-axis in the s-plane
Because the Laplace transform consists of s for which x(t)e-st
converges, which only depends on Re{s} = s
Property 2: For rational Laplace transforms, the ROC does
not contain any poles
Because X(s) is infinite at a pole, the integral must not
converge.
Property 3: if x(t) is finite duration and is absolutely
integrable then the ROC is the entire s-plane.
Because x(t) is magnitude bounded, multiplication by any
exponential over a finite interval is also bounded.
Therefore the Laplace integral converges for any s.
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Inverse Laplace Transform
The Laplace transform of a signal x(t) is:

X (s  jw )  F{x(t )e }   x(t )e st e jwt dt
st

We can invert this relationship using the inverse Fourier
transform
1 
st
1
jwt
x(t )e  F { X (s  jw )} 
X
(
s

j
w
)
e
dw



2
Multiplying both sides by est:
1
x(t ) 
2



X (s  jw )e (s  jw )t dw
Therefore, we can recover x(t) from X(s), where the real
component is fixed and we integrate over the imaginary
part, noting that ds = jdw
1 s  j
st
x(t ) 
X
(
s
)
e
ds

s

j

2j
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Inverse Laplace Transform Interpretation
1 s  j
st
x(t ) 
X
(
s
)
e
ds

s

j

2j
Just about all real-valued signals, x(t), can be represented as
a weighted, X(s), integral of complex exponentials, est.
The contour of integration is a straight line (in the complex
plane) from s-j to s+j (we won’t be explicitly
evaluating this, just spotting known transformations)
We can choose any s for this integration line, as long as the
integral converges
For the class of rational Laplace transforms, we can
express X(s) as partial fractions to determine the inverse
Fourier transform.
M
Ai
X ( s)  
i 1 s  ai
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L1{Ai /(s  ai )}
Ai eait u(t ) Re{s}  ai
 Ai eait u(t ) Re{s}  ai
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Example 1: Inverting the Laplace Transform
Consider when
X ( s) 
1
( s  1)(s  2)
( s)  1
Like the inverse Fourier transform, expand as partial
fractions
X ( s) 
1
1
1
A
B




(s  1)(s  2) ( s  1) ( s  2) ( s  1) ( s  2)
Pole-zero plots and ROC for combined & individual terms
x
-2
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x
-1
1
e u (t ) 
, Re{s}  1
s 1
L
1
 2t
e u (t ) 
, Re{s}  2
s2
L
1
t
 2t
x(t )  (e  e )u (t ) 
,
( s  1)(s  2)
t
Im
Re
L
Re{s}  1
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Example 2
Consider when
X ( s) 
1
( s  1)(s  2)
Re{s}  2
Like the inverse Fourier transform, expand as partial
fractions
1
1
1
A
B
X ( s) 




(s  1)(s  2) ( s  1) ( s  2) ( s  1) ( s  2)
Pole-zero plots and ROC for combined & individual terms
Im
-2
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x
x
-1
Re
1
, Re{s}  1
s 1
L
1
 2t
 e u (t ) 
, Re{s}  2
s2
L
1
t
 2t
x(t )  (e  e )u (t ) 
,
( s  1)(s  2)
t
L
 e u (t ) 
Re{s}  2
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Lecture 13: Summary
For many signals that are made up of a linear combination of
complex exponentials and CT LTI systems that are
described by differential equations, the Laplace transform is
rational, i.e. it is a ratio of polynomials in s: N(s)/D(s)
The roots of N(s) and D(s) are known as the zeros and poles
of the transfer function, respectively.
The Region of Convergence does not contain any poles
The inverse Laplace transform is given by
1 s  j
st
x(t ) 
X
(
s
)
e
ds

s

j

2j
It is usually calculated by expressing the Laplace transform as
partial fractions, and then spotting known relationships
(rather than directly evaluating the inverse transform)
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Questions
Theory
L
SaS, O&W, Q9.9, 9.22. Also, prove  (t )  1
Matlab
Verify Q9.9 in Matlab via
>>
>>
>>
>>
>>
syms s
y = ilaplace(2*(s+2)/(s^2+7*s+12))
t = 0:0.05:2;
y1 = subs(y);
plot(t,y1);
Do the same for the other examples in the help section for
ilaplace. Note that in Matlab dirac(t) is the
impulse/delta function (t) and heaviside(t) is the step
function u(t)
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